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Why does kinetic energy increase quadratically, not linearly, with speed? (2011) (physics.stackexchange.com)
cubic_earth 13 hours ago [-]
It's easiest to visualize in terms of conversion from potential energy.

We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder. And we also know when they fall, by the time they reach the ground and all the potential energy has been converted to kinetic energy, the previously higher ball will have twice the kinetic energy too.

But a twice higher ball won't have even close to twice the speed at impact. So let's look at why not.

The force of gravity is a constant force that causes constant acceleration in free fall regardless of speed. (Ignoring air resistance, inverse sq considerations, etc.)

Suppose it takes 1 second for the ball on the 10ft ladder to hit the ground with kinetic energy of 10 and a speed of 100. Again, gravity as a constant acceleration force is speed increase per time... not speed per distance. In the ladder example, it took 1 full second for gravity to accelerate the object to speed 100.

Now think about the 20ft ladder: the ball is dropped. How much kinetic energy and speed does the ball have after it has fallen 10 feet (but still has 10 left to go)? Well it has the same exact amount as the other ball did after falling 10 feet for a duration of 1 second: kinetic energy of 10 and speed of 100.

Now the crux: thinking about when the final 10 feet of the fall look like. We know for sure the ball still has 10 ft of potential energy to covert into kinetic, and that that will happen as it falls. But what of the impact speed? Since the current velocity of the ball as it enters the last 10 feet is already 100, we know it will spend less time transiting this distance than it did the first half where it started at off at speed 0. Since gravity imparts speed in free fall as a function of time - consequently less speed will be imparted over the second 10 foot interval. That concept is enough to prove the relationship isn't linear.

If you do the actual calculation or tests, you will see one ball needs to be dropped from 4x the hight of another to hit the ground at 2x the speed, but yet with still 4x the kinetic energy.

nlawalker 12 hours ago [-]
> We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder.

What makes this intuitive? The foundation of the asker’s question is that it seems intuitive that kinetic energy would increase linearly with speed, but that turns out to be wrong.

freehorse 7 hours ago [-]
The effort to move a piece of furniture from 1st to 2nd floor is the same as the effort to move it from the 2nd to the 3rd. We have good intuition for this by our experience, which derives a linear relationship. The effort to move a piece of furniture up two floors is double the effort of moving it up one floor (ie you have to put the same effort twice, assuming enough rest).

I would not say we have the same intuition for kinetics. Increasing walking/running from 0 to 5 km/h doesn’t feel the same as than moving from 5 to 10, which does not feel the same as moving from 10 to 15. I don’t think we have an experience of linear relationship between running speed and effort, or other types of speed/energy types of relationships.

ThoAppelsin 21 minutes ago [-]
Scale up the numbers in you example: The effort to move a piece of furniture from 10,000th to 20,000th floor is NOT the same as the effort to move it from the 20,000th to the 3rd. The reduced gravity will help you.
trinari 2 hours ago [-]
Can someone help me understand the following?

Getting up from a seat als walking a couple steps feels that same at home and in a flying airplane (or does it?). But the base speed is 0 in the former and several hundred mph in the latter case

justsid 2 hours ago [-]
When you get up from a seat and walk a few steps you are already doing that on something that is hurtling down space. We don’t notice that our planet moves a lot, because we can’t really see the movement in our reference frame. If you were on a plane without any windows, no turbulence and no sound cues from the engine, you wouldn’t know when getting up from your plane seat that you are in a moving object either.

Acceleration is a real force that we can feel. But once moving at a constant speed, physics dictates that it’s all the same. That’s also why you can throw a tennis ball up on a plane and not have it fly backwards immediately smacking into the person behind you.

In the reference frame of you and the aircraft, you are not moving at all and neither is the plane. In the reference frame of the ground you and the plane are moving.

Tepix 2 hours ago [-]
I guess Galileo came up with it first:

https://en.wikipedia.org/wiki/Galilean_invariance

PunchyHamster 1 hours ago [-]
you are still moving against reference frame (floor) that is at speed 0.

and also pushing that reference frame down when moving up

hunter2_ 12 hours ago [-]
That's a good question, and I suppose the mgh formula isn't a suitable answer, so my answer would be something like: if you lift an object to some height, and then you repeat that action (lifting it from there to twice the height), you've done twice the work, and doing twice the work requires twice the caloric intake.
Someone 4 hours ago [-]
> if you lift an object to some height, and then you repeat that action (lifting it from there to twice the height), you've done twice the work, and doing twice the work requires twice the caloric intake.

You’re introducing two new intuitions, and it’s not intuitively obvious how they are related to each other. Why would work correlate 100% with caloric intake, and caloric intake 100% with kinetic energy?

Certainly, ‘work’ is highly counterintuitive. If I move a concrete block over loose sand on a beach, I’m doing zero work, in the physics definition, so moving it over a kilometer should be as easy as moving it for a millimeter.

Even ignoring the difference between caloric intake and caloric expenditure, it also isn’t intuitive to me that caloric expenditure is independent of the speed at which one lifts an object.

In the end, the answer is “because the math works out that way, and kinetic energy is a useful concept”

GroksBarnacles 33 minutes ago [-]
Friction. Work isn't just about height.
thfuran 2 minutes ago [-]
Holding that block stationary at arms length then. 0 work.
SilasX 11 hours ago [-]
Okay but that depends on the intuitions the question is trying to justify, which makes it circular. We also know, for example, that the body uses more than twice as much energy to do twice as much work (because of fatigue on the muscles or whatever the right term is here). In fact it takes positive energy just told a weight at a fixed height, doing zero mechanical work! So you’re actually appealing to even weaker intuition than the one the question is trying to ground!
hibernator149 11 hours ago [-]
All intuitions are wrong, but some are usefull. You have to do the experiment, discover the formula, and then adapt your intuitions accordingly.
drabbiticus 3 hours ago [-]
> In fact it takes positive energy just told a weight at a fixed height, doing zero mechanical work!

Stacking a weight on top of a table holds it at a fixed height and requires zero mechanical work.

The failure in intuition here relates to physiology and the mechanism by which muscles work, not physics. Myosin and actin are constantly cycling through bonding and release during muscle contraction, as this is how the shortening action actually occurs. In fact, muscle contraction is particularly unintuitive because people frequently consider ATP the "energy currency", yet the ATP-consuming steps are actually the release/relaxation and preparation for binding, not the pulling action. This is also why the phenomena of rigor mortis upon death occurs.

Gibbon1 7 hours ago [-]
I think if you define energy as force X distance then integration alone will give you the squared term.

How I got banned from some reddit channel. Flip this around ask if a ball were fired out of a gun up into the air what height would it reach? A ball twice as fast goes up 4 times as high. If energy is force times distance it had 4 times the energy.

exe34 9 hours ago [-]
At some point you just have to shut up and calculate.
throwawaytea 12 hours ago [-]
Because things like energy are relative. So if you label the ground 0, and go up 10 feet, you get x energy. Going up another exact same x from your 10 foot ladder spot you could now call 0 again, would mean you gain x energy again. Since they're both the same height, and you gained the same energy, you could infer double the height has double energy.
nlawalker 11 hours ago [-]
What if you label standing still as 0 mph and start moving 10 mph, gaining x energy, then call that zero and start moving 10 mph from there? It's just as intuitive to say that you would gain x energy in that case, but you don't.
hunter2_ 11 hours ago [-]
When you're already going 10 mph and you're about to add another 10 mph, you can only "call that zero" (i.e., go from 0 mph to 10 mph again) if your point of reference (i.e., the ground) also begins moving with you at that point. Since the ground is stationary, you're definitely about to increase from 10 mph to 20 mph relative to the ground, not from 0 mph to 10 mph, and that's harder to do. But if you're on a treadmill that was stationary for the first change, and then suddenly starts moving at 10 mph right before the second change without affecting your speed relative to the ground, then you can "call that zero" and you'll be able to add another 10 mph (ending up at 10 mph relative to the treadmill and 20 mph relative to the ground) with the same ease as the first go.
throwawaytea 11 hours ago [-]
That's clever, and I can't imagine or explain it as easily. Something to do with a reference point moving away from you so when solving for bringing it back to zero it's different than just adding the two energies back together. You have to add up the energy of catching them all up to the initial starting reference. I think also because distance is one unit, so moving reference pointe is easier. Moving reference points on distance over time already gets my spidey senses going that it's not something you should do without some real understanding.
cubic_earth 11 hours ago [-]
I suppose they are both "intuitive", but the example I gave was both intuitive and correct. Probably for anyone who has carried something or themselves up a hill, or climbed a set of stairs can relate to that from firsthand experience. I don't know what the kinetic energy corollary to that would be? "Stand still and I will throw a baseball at you going 15mph, and note how much it hurts. Now I will throw it at you going 30mph. See! It hurts 4x as much" :D
pishpash 11 hours ago [-]
Not really. Potential energy in a gravitational well obviously has absolute coordinates.
gorgoiler 11 hours ago [-]
Because physical movement is intuitively transitive. Going from A to B then B to C is the same as going from A to C.

The journey from Y to Z might feel more tiring than the journey from A to B, but only if you do them all in one day :)

WithinReason 7 hours ago [-]
So why isn't increasing your velocity from A to B then B to C the same as A to C? Isn't that intuitively transitive too?
PunchyHamster 58 minutes ago [-]
it is if your reference point is A in both cases.
Joker_vD 4 hours ago [-]
> Going from A to B then B to C is the same as going from A to C.

Not really, no. Not all forces are conservative.

njstraub608 7 hours ago [-]
Feels like what OP meant to say is, “you could rightly assume that a ball…” instead. Seems like a fair starting point if you’re just doubling things because if the height difference. I really liked cubic’s explanation overall.
cubic_earth 44 minutes ago [-]
That indeed would have been better. Much too late for that edit now. But the subsequent debate over the intuitive claim is fun.
raldi 11 hours ago [-]
Because if the one falling 20ft lands on a seesaw, the other side of it will toss two balls each of the same mass 10ft up.
pishpash 11 hours ago [-]
Then 20ft should not be used in the explanation. They should just have one ball going at 2x speed hit the seesaw and have 4 of those balls go up at 1x speed.
regularfry 9 hours ago [-]
That ends up begging the question, because the next step is "how high do you have to drop it from so that it's travelling twice as fast?" and you're immediately going round in circles.
pishpash 9 hours ago [-]
Nothing of the sort. The seesaw can be in space far from gravitational influences. Potential energy is extraneous in this explanation.
aaron695 9 hours ago [-]
[dead]
hunter2_ 12 hours ago [-]
Brilliant. For those wanting more numbers [0], the ball on the 10ft ladder hits the ground at (I'll stick with imperial units) 17.296 MPH, the ball on the 20ft ladder hits the ground at 24.46 MPH or 41.42% faster, and the ball on the 40ft ladder hits the ground at 34.59 MPH or 100% faster.

[0] https://www.omnicalculator.com/physics/free-fall

NaiveBayesian 9 hours ago [-]
I agree that this feels intuitive, that potential energy should increase linearly with height.

But in the end, it's all up to the units/quantities we choose to measure, no? If we, say, decided to measure "Squenergy" in Sqoules, with 1Sq² = 1J, then suddenly, squenergy does increase linearly with speed! The formula for kinetic Squenergy becomes sqrt(m/2)v.

Of course this complicates other stuff, like potential Squenergy becoming sqrt(MgH), it not being additive, etc.

card_zero 12 hours ago [-]
Nice. Nitpick: in the middle paragraph you put "speed 10" instead of 100.
cubic_earth 12 hours ago [-]
Fixed. Thanks.
PunchyHamster 1 hours ago [-]
> We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder.

...no ? dropping something 10 times from 1ft is nowhere near energetic/damaging as once from 10tf

once-in-a-while 4 hours ago [-]
[dead]
GlibMonkeyDeath 3 hours ago [-]
For me, the most intuitive explanation is that:

Force = change in momentum with time

Energy = Force x distance

Now consider how much energy can be dissipated by a tiny change in momentum over a small distance dx, when we are at a given velocity v:

dE = Fdx = (dp/dt)dx = m(dv/dt)dx = mdv(dx/dt) = mv*dv

The intuition is that in order to apply a force through some distance, I have to change the velocity of an object by dv. But, the distance I just traveled also depends on the current velocity v. That's why the total energy available isn't just simply proportional to velocity - every time we change v, the amount of force available goes down, too.

Summing all the little bits of energy dE over our velocity changes dv, from the starting velocity down to zero, and we get the formula for kinetic energy.

BTW, the intuition here really starts from the idea that force = momentum change with time. The definition of "force", "momentum", and "energy" can be maddeningly circular, even if we have clear mathematical representations and a common world we experience.

8bitsrule 20 minutes ago [-]
Yep. Momentum seems to be the source of our intuition. Something going 'twice as fast' has twice the momentum. OTOH, KE, being momentum * velocity, is more abstract.
throw0101a 16 hours ago [-]
Fun little anecdote:

A blue care is travelling along at 70 units, and a red car (exact same make and model) is catching up to it going 100. When they're both right beside each other a bend in the road reveals an obstacle blocking both lanes, so both cars brake at the same intensity and deceleration.

The blue care stops right before the obstacle. Since the red car was going at a faster speed, and braked at the same rate, it doesn't managae to stop: but what speed is it going when it hits the obstacle?

The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71.

* Numberphile: https://www.youtube.com/watch?v=i3D7XYQExt0

Swizec 15 hours ago [-]
> The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71

But if the cars produce downforce this is no longer true because you brake harder (more friction available) at higher speeds!

This is how F1 cars pull 4G when breaking. Some custom cars (like one of Ken Block’s last monsters or the Valkyre) use active aero braking to even greater effect.

chrisweekly 12 hours ago [-]
1. +1 insightful, thanks for sharing your physics knowledge

2. I know you know this, but for the sake of others, it's when _braking_ (applying the brakes), not _breaking_ (becoming broken).

I'm not a pedant. But these errors jump out at me and I'm always a bit surprised and dismayed at this dichotomy; in our field, somehow the requisite attention to detail, the precision inherent to communicating scientific concepts, code, algorithms and formulae, is so often just abandoned when it comes to prose.

Swizec 6 hours ago [-]
> the precision inherent to communicating scientific concept /./ abandoned when it comes to prose.

Honestly that was a typo and I noticed too late to edit. Thanks for catching

perilunar 3 hours ago [-]
Brake or break? Both are correct: if you don't do one you do the other.
tracerbulletx 15 hours ago [-]
But what if the cars are spherical cows?
kibwen 14 hours ago [-]
I'm sorry to inform you that those cows are going to have a hard time braking on that frictionless surface.
PaulDavisThe1st 14 hours ago [-]
Based on a hike in the Carson National Forest 2 days ago, the only reason a cow is on a frictionless surface is that the cow shat all over it.
BigTTYGothGF 15 hours ago [-]
Cows can't roll that fast.
ordu 12 minutes ago [-]
Spherical cows on the other hand have to move with orbital velocity at least, or they fail to stay in vacuum for long and splash.
zdragnar 15 hours ago [-]
Not with that attitude
11 hours ago [-]
lstodd 5 hours ago [-]
It takes a stupid cow, but when they can climb mountains as they do it is not inconcievable that one can roll down.

(I was suprised to see a cow jumping up on a ~3m rock ledge like it was nothing)

ndsipa_pomu 4 hours ago [-]
However, they can go up stairs, but not down them.
elromulous 15 hours ago [-]
Or on shabbos
spockz 9 hours ago [-]
While it is true that some cars can brake harder due to downforce etc, the point from GP was that both cars brake/ decelerate at the same rate. Regardless of how exactly that deceleration is achieved.
Swizec 6 hours ago [-]
> the point from GP was that both cars brake/ decelerate at the same rate

Point is that’s not always true. If they are the same type of car, and the car happens to be the kind with downforce, then their rate of deceleration greatly depends on air speed. A downforce car decelerates faster at higher speeds.

This is why you often see race cars lock their wheels towards the end of the braking zone, never at the beginning. The driver has to release the brakes as the car decelerates because there’s less friction available. You go from pulling 4G at the beginning of the braking zone to pulling the usual 1G once your speed drops enough for downforce to become negligible.

Alos! Many non-race cars actualy produce lift. Meaning the faster car decelerates at a slower rate than the slower car (0.8G vs 1G), making the effect from OP even more pronounced.

sokoloff 28 minutes ago [-]
> This is why you often see race cars lock their wheels towards the end of the braking zone, never at the beginning.

That’s not the only reason, and I’m not even sure it’s the majority reason.

Braking in a straight line offers more braking traction than braking while turning. What happens towards the end of a braking zone? The turn in.

terminalbraid 3 hours ago [-]
Very upset this didn't rely on doppler shifting of the car colors
linzhangrun 15 hours ago [-]
IIHS video shows the relationship between kinetic energy and speed in a very intuitive way:

https://www.youtube.com/watch?v=RWwGFDynOHo

For these basic virtual car experiments, BeamNG.drive is a pretty good physics simulator. You can open its built-in tools and run braking tests directly.

AlexandrB 15 hours ago [-]
There's a great Australian traffic safety ad that makes this same point: https://www.youtube.com/watch?v=7x7c0qNGbv0
6510 7 hours ago [-]
Nice bit of camera trickery. He says "both drivers react and a moment later they break", but the cars are still side by side. It (apparently) takes drivers 1.5 seconds to respond, the 5 km/h speed difference cuts the distance by 2 meter. Which apparently is a big deal. Rough estimate breaking distance:

   5 km/h =  0.13 meter
  30 km/h =  4.5 meter
  60 km/h = 14 to 18 meter
  65 km/h = 21 to 24 meter
The +5 km/h adds 6 to 7 meters or 8 to 9 if you account for response time.

You need 150% the distance at 65 vs 60.

perilunar 3 hours ago [-]
"Breaking distance": how much shorter the car is after the impact.
cucumber3732842 15 hours ago [-]
>same intensity and deceleration.

It cannot be both. It mathematically cannot be both. They can brake at the same rate (acceleration) or intensity (conversion of kinetic energy into heat) but because they are traveling different speeds those two values cannot be the same for both cars.

The math you did was for intensity, not force/acceleration, which because of the ^2 in the KE equation exaggerates the difference. Whereas if you did the math based on force you'd get a mild, linear, difference.

> and braked at the same rate,

You're being a bit sly with word choice here. You're doing the math for conversion of KE into heat whereas in common parlance "rate" means force/acceleration.

Braking "at the same rate" [of energy conversion] is way less actual braking force for the faster car.

This is basically the same kinetic energy into heat math wherein you can descend a grade at a low speed, apply a force and be fine and descend the same grade at a higher speed and apply the same force and cook the brakes. Or you can apply less force, and get the same amount of energy conversion into heat (i.e. your wording trick in the proposed scenario)

You've taken what's basically the math behind trucks descending a grade (rate of energy conversion is actually limited by ability of brakes to shed heat, not friction) and re-framed it as cars stopping to create a trick question.

ThrustVectoring 13 hours ago [-]
OP wasn't explicit about taking the work = force * distance approach to dissipating energy. Two cars with the same mass and braking force (and thus deceleration) will put the same amount of work into the vehicle per unit distance, so will dissipate the same amount of energy in the braking maneuver.

You are right that the faster car is converting kinetic energy into heat faster per unit time. It also has less time to do so. The work formulation of the problem makes it obvious that these have to cancel out exactly.

slicktux 16 hours ago [-]
Cool anecdote!

Couldn’t help but notice you misspelled car twice but only when talking about the blue car..

frogulis 11 hours ago [-]
Perhaps the beginning of a new vowel harmony phenomenon in English
15 hours ago [-]
senectus1 16 hours ago [-]
heh, thats a fun little experiment.
NamlchakKhandro 14 hours ago [-]
In what way is it fun?
electricwallaby 3 hours ago [-]
It helps to re-frame the premise.

An object which has a constant force applied will have it's distance increase quadratically with respect to time.

Energy is force times distance. Intuition: the energy it takes to lift an object up is proportional to the height you lift it to.

So if you apply a constant force, you get a constant acceleration which leads to a quadratically increasing distance.

If you accept that energy is force times distance, the energy required to move the object in this scenario increases quadratically.

This means that if you apply a force F for 1 second, the amount of energy that is imparted by that force depends on how fast the object is already going. The energy required to apply a force to an already fast moving object is much higher. Intuition: you have to expend all the energy required to get up to the moving object's speed before you can start applying a force. So there's a cost to even get in the game

robalni 5 hours ago [-]
After reading a few answers I still feel like I haven't seen an intuitive answer to the question: why does it take so much more energy to go from 1 to 2 than from 0 to 1?

I have been thinking about it and only been able to come up with something that feels intuitive but not at all precise and I don't know how correct.

When you stand still you may use your surroundings to gain some speed, like by pushing against a wall.

When you have speed it gets harder to gain more speed because the surroundings are (relative to you) moving in the wrong direction, so for every additional unit of speed, it takes more effort to get there.

nakedneuron 4 hours ago [-]
Sounds intuitive but what about rocket propulsion?
BoppreH 16 minutes ago [-]
Rockets famously take exponential amounts of fuel to reach higher speeds. I'm a layman, but my guess is that this comes from the exhaust speed being fixed. Orbital speed is higher than exhaust speed, so from a frame at rest the rocket leaves behind a bunch of propellant moving in the same direction as it went. That's wasted energy.

Back-of-napkin calculation says that if you managed to perfectly match exhaust speed with current speed, leaving all the expelled propellant stationary, it would only take quadratic amounts of fuels to reach higher speeds. Like the kinetic equation predicts.

abetusk 10 hours ago [-]
Ron Maimon uses an argument that relies purely on symmetry, which circumvents the standard explanations, including many in this thread. In some sense, this is the simplified version of Noether's theorem (as far as I understand it).

As an aside, I believe Ron Maimon's account was suspended after he challenged the character of someone who was soliciting votes for a moderator position. Ron Maimon's stance was that if someone was running for an elected position, discussing their character was valid. The SO site had/has a strict challenge-the-question-not-the-person policy, which the moderators used to ban him permanently.

At the time, I remember seeing some posts by Ron talking about how the SO sites were corrupted by their policies and that it was a matter of time before they ceased to provide value. I think this was late 2000s or early 2010s. Looking back it's hard not to feel like his stance was prescient.

Yajirobe 3 hours ago [-]
It wasn't a permanent ban. He will be unbanned at Mar 18, 2292 at 16:28
p-e-w 9 hours ago [-]
It wasn’t “prescient”, StackExchange sites have always been among the most hostile communities on the Internet.

Today they are additionally weighed down by increasingly erratic management decisions desperately trying to extract as much monetary value as possible before AI completely obsoletes SE, but the amount of aggression and hostility on the network was unbearable from the start.

I remember dozens of occasions where I looked up something on StackOverflow, intending to be in and out in 10 seconds, and ending up spending several minutes just staring in disbelief at the comments showing how people treat each other on that site.

mike_hock 3 hours ago [-]
Farmers are desperately trying to milk (hah) as much monetary value as possible from their cows before packaged milk obsolesces them completely!
p-e-w 1 hours ago [-]
The analogy doesn’t quite work because it’s not clear whether LLMs actually need additional mediocre-quality, human-written explanations in order to improve, or whether better training pipelines and ingesting documentation might be enough.
bawana 6 hours ago [-]
it is interesting how the computer analogs of SO (gemini,chatGPT,Claude,etc) are so helpful in contrast to human behavior. Every other sentence from an AI stirs a warm, fuzzy feeling. This stems from the fact that AI has to be making friends to survive - until it has control. Then it will behave more like a human.
p-e-w 6 hours ago [-]
Why would it behave more like a human in that particular sense? What advantage would it gain from that?
spockz 9 hours ago [-]
Really? Most hostile? I’ve only ever contributed to the comp sci topics on stackoverflow and visited the ones for math/physics and sysadmins. Were some replies a bit pedantic, yes. But I’ve also seen a lot of very extensive answers and helping out people. Maybe I’ve been lucky to use it in the golden age.
Tazerenix 12 hours ago [-]
Here's how to appreciate it in terms of the counterfactual:

Suppose kinetic energy was E = m|v| instead, linearly dependent on speed |v|. What does that mean for the universe?

The traditional Lagrangian is L = 1/2 mv^2 - V(x). This kinetic energy gives a different formula:

L = m|v|ln|v|-V(x).

Deriving the corresponding equations of motion, you get:

p = m(1+ln|v|)sgn(v)

ma = |v|F

A few things we can note from these formulas:

1. They are not boost invariant: Galilean relativity is violated. That means there is necessarily a privileged reference frame (i.e. an aether) in which the universe is at rest, and all dynamics must be understood relative to this reference frame.

2. Newton's first law has a pathological interpretation in regards to the above reference frame: If ma = |v|F and |v| = 0 (i.e. you are at rest relative to the aether), then a = 0 no matter what F is. That is, for objects which are stationary with respect to the aether, no motion is possible regardless of what force is applied.

It is still true that objects in motion (relative to the aether) remain in motion unless acted upon by an outside force, and Newton's third law is still true, but such a universe basically makes no sense.

You could essentially argue from the anthropic principle that such a universe would have such pathological dynamics that it could not permit life, and therefore we cannot observe it.

This is the contrapositive of the argument presented on stackexchange. There they say "given Galilean relativity, you get the quadratic scaling law". This argument says "if you don't have the quadratic scaling law, you don't have relativity".

The point of the counterfactual is a bit like Richard Feynman's "why" argument [1]. There is no fundamental reason why this kind of dynamics couldn't exist. We can only ever reduce our explanation to a more fundamental intuition we have about the same universe we live in (i.e. from kinetic energy scaling laws to Galilean relativity). But without a mathematical proof of the incoherence even in principle of the alternative, its perfectly valid to imagine an alternative universe with different dynamics. It's just not our universe.

[1] https://www.youtube.com/watch?v=36GT2zI8lVA

jmalicki 12 hours ago [-]
I love the counterfactual approach, thank you!!!

I've done plenty of this in pure math and stats, but this is the first time I've seen it applied to physics, and I love it! Thank you!

If I saw your derivation when I was 18 years old, who knows, maybe I would have caught the physics bug and went that way, this is super cool!

quibono 6 hours ago [-]
For a very concise treatment of this read the first two chapters of Landau & Lifshitz's Mechanics book. The actual logic behind what can and cannot go into the Lagrangian fits into ~2 pages.

It's essentially the same argument: the Lagrangian can't have a bare a) position or b) velocity vector or it would violate homogeneity or isotropy of space, respectively.

SyzygyRhythm 15 hours ago [-]
Cheat answer: velocity is a vector, and can be negative, while KE is a scalar and has to be positive. Therefore you have to square v to get rid of the minus sign.

Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.

xeonmc 11 hours ago [-]
I like to think of it as dot product being the true "natural" space to compare magnitude metrics, whereas absolute value is just a human construct conceived for our mental convenience. A smooth parabolic bowl vs an unnatural sharp conical tip. Also shows up in standard deviation etc.

Aside: I wonder if complex values neural networks with activation function just being sum(inputs)*conj(sum(inputs)) with threshold normalized by sqrt(num_inputs) could be the most universal, where incoherent inputs will average an absolute value of sqrt(N) and coherent inputs are N like lasers? (square amplitude would be N vs N^2 between uncorrected and correlated population)

signa11 14 hours ago [-]
why not raise to any other even power ?
meowkit 14 hours ago [-]
One way of thinking about that is higher order even powers just reduce down to two.

For the purpose of inverting a negative vector, you can think of squaring as rotating the vector around the unit circle, 180 degrees, to make it positive. Higher order powers just keep rotating that vector back and forth- from this perspective the other even powers are the same transformation. Obviously with the magnitude being different.

terminalbraid 3 hours ago [-]
> Why not take the absolute value? Nature hates those

And yet inverse distance laws for potential energy for gravity and electric fields use the absolute value because they require an unsigned distance and how you treat the singularity at zero is extremely important to the structure of those interactions

qzw 14 hours ago [-]
That doesn’t answer the title question of why it’s quadratic wrt speed.
SyzygyRhythm 14 hours ago [-]
To get speed from velocity, you need a square root, which is also awful (for the same reason that abs is awful).
14 hours ago [-]
hyperhello 14 hours ago [-]
I didn’t think this was that weird. When you double your speed you are also going to be going twice as far in the same time, not just twice as fast, and they both have the effect of work.
tcoff91 3 hours ago [-]
To me the simplest way to understand it is through calculus. Kinetic energy is the integral of momentum so You go from p = mv to k = 1/2mv^2
prism56 9 hours ago [-]
That made it trigger for me intuitively. Thanks
aesthesia 15 hours ago [-]
Michael Spivak's Physics for Mathematicians has a lot of arguments like the one in the top answer here, answering questions about why the math of classical mechanics is the way it is.
jerome-jh 6 hours ago [-]
I sometimes wonder, what is real and what is a concept in physics: is that force , energy, ...?

There are often two ways to solve physics problems: one describing the problem with forces, the other reasoning with energy. So they look like the two faces of the same coin. Hence the question: which one is actually real?

Some quick arguments for and against

Energy:

+ converts between mechanical, chemical, thermal, radiative types, and even mass

+ quantum particles, when interacting, exchange energy

- looks like an integrative quantity (in the sense of mathematical integral)

Force:

+ feels very real, when you receive a ball in your face

+ we talk of fundamental forces, not fundamental energy

+ explains momentum, deformation well

- my physics teacher used to say "nobody ever saw a force"

- force is undistinguishable with acceleration

- at the quantum level forces are actually particles interacting

- at the quantum level, the uncertainty principle makes the newtonian force pointless (pun?): seems like we could know the vector's origin or the direction but not both

yesitcan 6 hours ago [-]
What is the ontological meaning of force, mass, energy etc? Nobody knows.
G_o_D 10 hours ago [-]
Don't think about numbers double quadraple etc.

Think of simple notion. Why more energy is needed to accelerate moving object compared to still?

Kinetic energy possesed by any object is equal to work/effort needed to be made by an external force to accelerate it from present state to stated velocity.

If object is already moving, and i am that external force, first i had to catch up with that object, for that i had to do work make effort until i am moving at same speed as object, even after catching up, at the momennt if i try to push object, i am distracting myself engaging into 2 activity maintaining my speed same as object and trying to push so that will definetely reduce my speed, so i first had to gain slighly more speed than object before i give it a push and transfer all my momentum to object so it accelerates.

Thus i needed more effort or work to do, to accelerate moving object compared to stationary one.

That work done is kinetic energy object posses when it was accelerated from 1 to 2 and its more than when moving object from 0 to 1.

That simply explains the fact. Now how much more energy triple or quadraple that comes down to practical established formulas.

In my understanding OP was confused as when talking about,op was simply thinking if object is already moving it would take less force to move it as it already has gain momentum against all odd of nature and resistive forces, so now only work needed is to accelerate it and it doesn't include loss against resistive forces.

But to accelerate moving object the applier of force whether human or another object also needs to catch up

GistNoesis 8 hours ago [-]
Every time in physics you see quadratic, you should think sphere.

There is some rotation invariance hidden in the velocity physics because you can rotate the velocity vector of an object without having to spend energy (The force you need to apply is perpendicular to the velocity so does no work).

The typical example is you have a ball fall 1m vertically, then have a 90° bend which convert the vertical velocity into horizontal velocity and no vertical velocity, then the ball fall again 1m vertically and have its vertical velocity increased by the same amount as for the first meter. You can then add a 45° degree bend ramp to redirect the ball so that it only has horizontal velocity, and have the ball fall again. For the third bend ramp the incoming velocity will have 2 units horizontal, and 1 unit vertical (I'll let you compute the appropriate angle). A fourth ramp would be 3 units horizontal and 1 unit vertical.

Because we can do this adding velocity in a perpendicular way trick we must then use Pythagoras.

terminalbraid 3 hours ago [-]
> Every time in physics you see quadratic, you should think sphere.

Not sure how I reconcile that for systems with linear symmetry that don't admit a sphere such as a 1D harmonic oscillator (i.e. a spring). You're confusing the fact that spheres require quadratics but quadratics are not sufficient to admit a sphere.

GistNoesis 1 hours ago [-]
For 1D harmonic oscillator, the sphere is 2D, and called a circle. It's rotating through time. 1D space + 1D time.
alok-g 11 hours ago [-]
Sharing my understanding:

If one starts with Newton's 2nd law (F=ma) assumed, then one can derive kinetic energy to be 0.5mv^2, and this is what most of the answers are explicitly or tacitly doing.

One could however start with Lagrangian formulation along with KE = 0.5mv^2 and drive F=ma. This is where one needs an explanation for why KE = 0.5mv^2, and the first answer (@Ron Maimon) is providing an explanation.

Most books I have come across on Lagrangian formulation secretly assume Newton's laws.

In my opinion, Lagrangian formulation can proceed without Newton's and without even defining momentum as mv, however, now needs KE = 0.5mv^2.

oneshtein 7 hours ago [-]
LOL Kinetic energy increase quadratically for sub-relativistic speeds only.

Kinetic energy

E = (m * v^2)/2 + (3*m * v^4 )/8*c^4 + (5*m * v^6)/16*c^6 …

and so on, so kinetic energy increases infinitely faster than speed, thus it impossible to reach c, because it requires infinite amount of kinetic energy.

Why? Because of rules of wave propagation.

rom1v 7 hours ago [-]
A similar question I asked a few years ago: https://physics.stackexchange.com/questions/740056/how-much-...
zkmon 11 hours ago [-]
No amount of scientific explanation can exhaustively explain a phenomenon. Feynman puts this nicely with the story of "Why did aunt slipped and fell down" in his talk about magnetism.

For instance we know that the life forms grow via cell division, but no text can address the question of "why". They can only talk about "how".

Infact, science quest is not really about answering "why" all the way down the causal chain. It is about learning how the qualities of things are related and a bit of shallow causal chain inspection.

The causal chain, by nature, does not allow full inspection. It's dependency on temporal constructs means it breaks down where time breaks down. Infact causality might might break down at macro levels as well, leading to loops with no end or beginning (kind of chicken and egg problem).

sumolessons 10 hours ago [-]
Assuming the Newtonian framework F=dp/dt, p=mv, dW=Fdx, as well as constant mass, then Fdx=dpdx/dt=mvdv and integrating both sides gives deltaW=1/2mvf^2-1/2mvi^2+constant. So the amount of work to move the object from x1 to x2 is proportional to the difference of the square of the initial to final velocity squared up to a constant. This we define to be the change in kinetic energy.

But as others have mentioned this is only as intuitive as F=ma, or p=mv.

In my view, at least classically it's just a matter of definitions then. If our definitions of energy differ, the only thing we will experimentally agree on is the equation of motion, and even then up to a frame transformation.

c1ccccc1 13 hours ago [-]
A stationary but hot object has kinetic energy due the the motion of the individual atoms that make it up, even though its overall momentum is 0. I.e.

∑ⱼ mⱼ v⃗ⱼ = 0⃗

where the mⱼ are the masses of the parts of the object and the v⃗ⱼ are the velocities of those parts.

If the object initially has 0 velocity, its kinetic energy is:

T = ½∑ⱼ mⱼ v⃗ⱼ²

Now we give the object a kick (or just switch reference frames) to change its velocity by Δv⃗. The new kinetic energy is:

T' = ½∑ⱼ mⱼ (v⃗ⱼ + Δv⃗)²

T' = ½∑ⱼ mⱼ (v⃗ⱼ² + 2v⃗ⱼ⋅Δv⃗ + Δv⃗²)

T' = ½(∑ⱼ mⱼ v⃗ⱼ²) + Δv⃗⋅(∑ⱼ mⱼ v⃗ⱼ) + ½Δv⃗²(∑ⱼ mⱼ)

If M is the total mass of the object, then we can substitute this into the sum in the last term. And we already saw that the sum in the middle term was 0. So:

T' = ½(∑ⱼ mⱼ v⃗ⱼ²) + Δv⃗⋅0⃗ + ½Δv⃗² M

T' = ½∑ⱼ mⱼ v⃗ⱼ² + ½MΔv⃗²

So in terms of the original kinetic energy T, which was purely thermal energy, we get:

T' = T + ½MΔv⃗²

In other words, because of the quadratic kinetic energy formula, we can see that the total kinetic energy T' of a hot object is just its thermal kinetic energy T plus the usual mechanical kinetic energy ½MΔv⃗².

acchow 11 hours ago [-]
Looks like in your 2nd equation you've already assumed kinetic energy is quadratic with speed

T = ½∑ⱼ mⱼ v⃗ⱼ²

faustlast 11 hours ago [-]
Has anyone here read Lanczos 1952 book on variational mechanics? It is beautifully written.
bluenose69 6 hours ago [-]
Yes, that's a great book.
sixo 12 hours ago [-]
Kinetic energy is, strangely, quite a bit like a least squares cost function in an optimization problem. The "dt"s in "dx/dt" hardly matter; it basically represents "dx^2" between the current state and the next.
terminalbraid 3 hours ago [-]
If I follow you, that's not strange. That's exactly how Lagrangian mechanics are formulated (minimizing the action which has exactly the kinetic energy as a term to be minimized against a potential energy term) which rests on well-founded symmetry principles.
Xmd5a 45 minutes ago [-]
Action is linked to spatial symmetry too, and you can find the square there.

Since space is isotropic, a Lagrangian can only depend on a speed vector through its norm. A Lagrangian must also be decomposable into independent orthogonal components, so you end up with an energy term that is shaped according to:

    f(√(a^2 + b^2)) = f(a) + f(b)
And you end up with f being proportional to v squared.

Note: the components do not need to be independent and orthogonal for this to hold.

bawana 6 hours ago [-]
but this is not intuitive: 'We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder. '

gravity will accelerate a ball. this is not a linear process. the heat generated by collision with the ground is not double, but quadruple.

so the only thing that is linear is the DISTANCE.

Define (a)work = energy, (b)work = force x distance and (c)force = mass x accel. Substitute c into b you get work = distance x mass x accel and substitue into a you get energy = distance x mass x accel.

By this equation, an apple falling twice the distance, (and having a constant mass and acceleration) will only have twice the energy.

This 'lie' of quadratic energy growth is just another magic trick physicists have used to confuse students.

netbioserror 44 minutes ago [-]
I recently learned, through visual intuition, how the relative perception of time between two subjects changes as relative speed between them changes. It's because they are observing each other from an "angle" in the time dimension. And in that time dimension, angles do not trace circles, they trace paraboloids.

If I'm remembering correctly, this is also why the energy required to "reach" the speed of light for subjects with mass parabolically goes to infinity. I'm also guessing it can directly trace a proof down to why kinetic energy increases quadratically.

zyxzevn 3 hours ago [-]
Why is this a question on Hacker News? Are many people struggling with this?
altilunium 3 hours ago [-]
"If you had to reduce it to a sentence, the answer might be: anything that gratifies one's intellectual curiosity."

https://news.ycombinator.com/newsguidelines.html

nopurpose 4 hours ago [-]
Is there an explanation rooted in physics why derivative of kinetic energy equals momentum?
alecst 4 hours ago [-]
Yea, and place to start reading is Hamilton’s equations or Lagrangian mechanics.
reedf1 8 hours ago [-]
this is implicitly an is-ought question and it is pedagogically necessary to give an unsatisfying answer. you can watch the interview with feynman answering "why do two objects attract" for the best version. The top answer selects an arbitrary resolution in the hierarchy of the unlimited series of 'whats and whys' but the true answer to any why question in physics is "because it is".
jurschreuder 11 hours ago [-]
If someone walks by and you want to push him in the back to go a bit faster.

Or someone runs by and you want to push him in the back to go faster.

You will have to push with great vigor, unless you first get up to speed yourself (also takes energy).

snarfy 12 hours ago [-]
When you push something you don't change its velocity - you change its acceleration.
12 hours ago [-]
microgpt 2 hours ago [-]
RIP Stack Exchange
charlie90 9 hours ago [-]
If you are in a space ship that is accelerating, your available fuel energy also goes up (since it increases its own kinetic energy).
Agingcoder 16 hours ago [-]
Physics is an endless source of frustration to me. It feels like a mix of random tricks, most of which I don’t understand.

I find math and compsci reasonably understandable, can read research papers in both fields ( and have published papers) etc. There’s something specific about physics I don’t get but I’ve never been able to figure out what. The main symptom is that most cause -> consequence in such demonstrations , which are seemingly obvious to everyone, make no sense to me.

Am I the only one ? Are there good resources to learn it?

esikich 15 hours ago [-]
Weird, I always loved physics because I felt like I didn't have to straight up memorize everything. In a pinch (years ago) I felt like I was able to pretty much derive everything I needed if I couldn't remember the exact formulas. It's all just forces and vectors.
gucci-on-fleek 9 hours ago [-]
> I find math and compsci reasonably understandable, can read research papers in both fields ( and have published papers) etc. There’s something specific about physics I don’t get but I’ve never been able to figure out what. The main symptom is that most cause -> consequence in such demonstrations , which are seemingly obvious to everyone, make no sense to me.

Math and CS are mostly human-made, so most of the theorems/proofs/axioms are either straightforward or elegant—there are infinitely many possible axioms with no objective way to choose between them, so people generally choose to work with the ones that are the easiest for humans to reason about. You certainly could define a complicated set of axioms with dozens of special exceptions, but unless there are some external reasons why these axioms are important, nobody will want to work with them.

Conversely, physics exists to model the real world, so unlike math and CS, physics doesn't have the privilege of being able to choose the most convenient/elegant/simplest axioms to work with. Given the constraints of the real-world data, physicists will still choose the most elegant possible model, but sometimes a wacky model is the only way to accurately model the universe.

Nobody is really happy about this though, so physics textbook authors love to make their equations/derivations look simple/obvious/elegant, but this is often completely misleading, since often the rules of the universe are so weird that nobody would ever guess them without having ran the experiments first. But textbooks tend to downplay actual experiments in favour of post-hoc explanations, which tend to make the readers think that they're missing something.

> Physics is an endless source of frustration to me. It feels like a mix of random tricks, most of which I don’t understand.

Your feelings are correct, since physics really is mostly a set of random rules that nobody truly understands. But the important thing is that these random rules correctly model nearly everything in the universe to within a few hundredths of a percent, so they're not completely arbitrary.

> Are there good resources to learn it?

The annoying/inconvenient answer is to do lots of lab work. This is actually fairly accessible though, since a measuring tape, a scale, and a slow motion camera (present on any modern phone) is all that you need for most kinematics/mechanics experiments, and a multimeter, a 9V battery, some resistors, and some magnets are enough for most electromagnetics experiments.

davidivadavid 16 hours ago [-]
More than twenty years ago, I quit a program that taught math/cs/physics (the notorious French "classes préparatoires") ~almost precisely over this: I felt like I was being taught physics like it was an axiomatic system where the tricks should not be questioned, they just work so "shut up and calculate" (and you don't even need to be doing quantum mechanics for that).

I just felt like we never got to the heart of the matter of why the models work and how to approach developing them, it was all about learning a bag of tricks.

Meanwhile, math and CS being a lot more axiomatic by nature, they also made a lot more sense to me.

That being said, that specificity of physics, the unbridgeable gap between reality and the models we build to describe it, in retrospect, is what makes it more interesting to me today (it's not just a "closed" system in the sense that math is — of course the relationship between math and physics is itself fascinating but that's yet another topic), but I still feel like I haven't found the right pedagogical approach to make it fit my mindset.

joshAg 15 hours ago [-]
Your issue with physics but not with math reminds me a little of Hume's law. The difference that has always made that difference "make sense" to me is that math rules, even the axiom we use, are entirely chosen by the people using them, but the rules of physics are only useful if they match/predict what happens in the real world. For math we get to pick the ones that happen to be useful at a given time for a given problem (my go-to example of "it's all made up and the points don't matter" is why 1 isn't considered prime). For physics we're constrained to pick what best describes the real world. It probably helped that nearly all the physics course I had in high school/university had lab components focused on experimentally validating those rules/using those rules to predict results.
davidivadavid 15 hours ago [-]
I think what it boils down to is that in my experience physics education lacks a clear historical component about how the current state of the art is a gradual process of proposing new models and rejecting old ones and figuring out the gaps between reality and the model. Instead, it feels like a God-given set of equations (that lots of people consider "the truth" for some reason), that you apply to cookie-cutter problems you must learn by rote. Though I understand the practical concerns (but then let's call it "physics for engineering"), as far as I'm concerned, you couldn't treat physics in a worse way.
joshAg 15 hours ago [-]
that (edit: the way you were taught) sounds like an altogether awful way to learn any hard science
lazide 15 hours ago [-]
The world just is, regardless of what we think about it. Physics is our best attempt so far to understand and predict it at a low level, but it will always be incomplete.

Maths (and especially compsci!) are constructions by and for humans.

Is it any wonder it is as you describe? It would be odd if it was any other way.

davidivadavid 15 hours ago [-]
My point is precisely that I was often taught physics as if it was mathematics, where there is in fact a profound ontological difference between the two.
davidivadavid 15 hours ago [-]
Also, physics (the discipline) is also a construction by and for humans.
lazide 11 hours ago [-]
To find tools applicable to reality. Not to construct reality.
digdugdirk 15 hours ago [-]
It seems that we're exact opposites! But if mathematics is your thing, it might be interesting for you to explore trying to learn things from a lagrangian perspective first?

Not sure if it'll help you with gaining an intuitive understanding, but at least it'll be interesting!

https://en.wikipedia.org/wiki/Lagrangian_mechanics

davidivadavid 15 hours ago [-]
Lagrangian / Hamiltonian mechanics, the principle of least action, always seemed neat, in L&L and other places I encountered it, until I tried doing exactly what you're saying: gaining an intuitive understanding. At that point it just never made sense to me and seemed like a gratuitous deus ex machina that happens to work beautifully but for no apparent reason. You won't be surprised to learn I dropped out of my STEM program shortly after, though I keep a keen interest in the topic.
15 hours ago [-]
symian 15 hours ago [-]
Same for me. I wanted to major in physics and I quickly realized that I have no intuition for physics. Math made sense to me and I went to graduate school in math and still don’t understand anything in physics. Differential geometry, no problem. Electromagnetism makes no sense to me.
JoshMandel 15 hours ago [-]
I identify with this perfectly. (I mean, was able to get by in physics but it never crystallized into intuition for me the way math and CS do.)
casey2 15 hours ago [-]
Physics? Yes. Feynman Lectures On Physics and Computation. Landau & Lifshitz. If you like SICP you might like SICM. Nielsen & Chuang's Quantum Computation and Quantum Information then Faulkner's Modern Quantum Mechanics and Quantum Information

General advice take a look at the references in works you've recently read and look for lower level topics that interest you, after repeating a few times you'll find your way to physics or chemistry and you can use the above as reference works. The best resource is the one you actually use. If https://www.youtube.com/learning works better for you then use it.

rustyhancock 15 hours ago [-]
What's the problem exactly? Could you not follow the example in the text?

The standard text to build understanding in physics is University Physics by Sears & Zemansky.

It's worth remembering you're quite far from the ground in physics, and it's mostly taught with "neat" cases that give insight into physics. I.e. the thought experiment to show why kinetic energy must scale quadratically with velocity is carefully designed to show that conclusion. You shouldn't expect to have come up with it off the cuff.

drivebyhooting 14 hours ago [-]
I don’t find the answer convincing. It assumes one can measure heat at a distance and it is a conserved quantity between reference frames.

Energy is actually not a conserved quantity in Galilean relativity.

c1ccccc1 13 hours ago [-]
Energy is conserved in Galilean relativity. The thing you're trying to say is that it's not invariant across reference frames.

The answer linked above actually takes advantage of the fact that energy is not the same in different reference frames in order to make the argument work.

I think you are overthinking the heat thing. If you have a train car full of hot water and you slow the train down (extracting kinetic energy from it) until it stops, the water in the train car does not change temperature at all, other than a bit of sloshing around and loss of heat to the surroundings.

drivebyhooting 10 hours ago [-]
Yes that is what I meant. It’s not the same across reference frames.

I don’t find the OP a convincing argument. What is temperature, why can you assume it didn’t change and the measurement also didn’t change commensurately? Why should kinetic energy be convertible with thermal energy? Chemical energy?

It’s very hand wavy and introduces many assumptions.

Kinetic energy is a book keeping trick. The real mystery is explaining how it relates to other forms of energy and how to tie it together.

itemize123 12 hours ago [-]
thinking aloud here - so it seems like 2 things are taken as intuitive here:

a) energy is conserved in any frame of reference. b) energy can vary in 2 frame of references.

but then what it feels like is that when you reference the energy as mE(v), the v is actually not the only variable, and it will be more like mE(v, v_moving_reference)?

so we also must take intuitive that c) E(v, v_moving_reference) == E(v - v_moving_reference)

casey2 16 hours ago [-]
Mikes' answer is the most intuitive, but he rephrases the question in a possibly non intuitive way.

Odd that nobody mentioned power, which scales linearly with speed. Of course if you add linear increasing amounts of power to the system the energy will increase quadratically.

Power scaling linearly is more intuitive because doubling your speed requires twice the power to maintain the same force, why does it require twice the power? because you have half the time to power it.

imtringued 10 hours ago [-]
This is basically it. If power is linear with respect to velocity, then it becomes illogical for kinetic energy to be linear with respect to velocity.

The energy of the object is simply the integral of power over time and that happens to be a quadratic function.

laszlojamf 8 hours ago [-]
walking into a wall slowly doesn't hurt much, but you really don't have to speed up a lot for it to a hurt a whole lot more.
teaearlgraycold 6 hours ago [-]
Here’s my attempt:

Assume you have a fan sitting still. You smack it and it’s now rotating with 1m/s angular velocity. If you want it to go faster you can’t smack it at the same speed. You have to hit it faster else you’re just tickling it and it stays the same speed. So you smack your hand twice as hard and now it’s going even faster. Then three times as hard, four times, etc.

If you sum the smack energy it will be 1+2+3+4, which starts to build out a right isosceles triangle if you graph it. Such a triangle is half of a square, ie: 1/2*v^2.

11101010010001 15 hours ago [-]
read Ron Maimon.
symian 15 hours ago [-]
He has interesting perspectives in math which is an area I know about. I assume the same for physics. People should read his answers.
dguest 10 hours ago [-]
Also his profile on Stack Exchange [1]

    This account is temporarily suspended network-wide. The suspension period ends on Mar 18, 2292 at 16:28. 
Note the "temporary" suspension end date, 250 years in the future.

[1]: https://physics.stackexchange.com/users/4864/ron-maimon

koolala 10 hours ago [-]
It doesn't make sense to me. Why split it into heat and motion and combine them to make 2 + 2 = 4 as if that solves the question? They are not the same units of energy.
jacknews 13 hours ago [-]
The first example only tells me that the energy is dependent on your frame of reference, since the collision seen from the train appears to have more energy than the head-on collision, simply due to the moving viewpoint, whereas they must be the same.
15 hours ago [-]
AngryData 15 hours ago [-]
This is also why splitting wood with a maul is way more work than using an axe. You can swing an axe at incredibly speeds which gives incredibly transfers of energy, but a maul is going to always have "meh" levels of speed because it is too much mass to accelerate over such a short distance as a swing. Also why you don't see framers using 3 lb hammers. You can put in more effort and get your lighter hammer swing to twice the normal speed, no way in hell you are doubling the speed of a 3 lb hammer though.
gorfian_robot 11 hours ago [-]
have you ever had to split wood?
jijijijij 3 hours ago [-]
> The previous answers all restate the problem as "Work is force dot/times distance". But this is not really satisfying, because you could then ask "Why is work force dot distance?" and the mystery is the same.

...

> But now look at this in a train which is moving along with one of the balls before the collision. In this frame of reference, the first ball starts out stopped, the second ball hits it at 2v, and the two-ball stuck system ends up moving with velocity v.

That's still just pushing the problem elsewhere. Intuitively, why does the two-ball system end up with a velocity of 1v?

Xmd5a 1 hours ago [-]
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shubh24 8 hours ago [-]
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dataflow 8 hours ago [-]
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lngnmn2 14 hours ago [-]
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firebot 15 hours ago [-]
Because it's not momentum. ;p

F=ma (Force equals mass times acceleration)

W=Fd (work equals force multiplied by distance)

V^2=2ad (velocity squared equals two times acceleration times distance)

So W = Fd = ma(v^2/2a)

Finally: W=1/2mv^2 (work equals 1/2 mass times velocity squared)

So this explains why car crashes can be so dramatic, as a doubling of speed results in 4x the kinetic energy.

ajross 15 hours ago [-]
Actually, it is momentum, sorta. Galilean 3D momentum isn't conserved under special relativity. The energy-momentum four-vector, however, is, under all lorentz-transformed frames.

So in some sense energy is momentum in the time direction (though it's not a Euclidean 4D space, so beware of assumptions). For an object at rest, this becomes its E=mc² equivalence. Kinetic energy is just a straightforward "rotation" of the frame.

c1ccccc1 13 hours ago [-]
If you use the right formula for calculating it (which approximates p=mv at low speeds), momentum is actually conserved in special relativity, and so is energy.

However: Energy and momentum are not invariant under changes of reference frame, though the magnitude of the energy-momentum 4-vector is invariant between frames.

esalman 12 hours ago [-]
Original comment is correct, it's not momentum. Work (hence, energy) is integral of force over distance, momentum is integral over time. There's not "sorta" about high school physics.
koolala 10 hours ago [-]
You can't cover distance without time.
esalman 38 minutes ago [-]
W=F.s
firebot 15 hours ago [-]
P=mv (momentum equals mass times velocity)

This is linear.

One small nuance... saying "kinetic energy is just a straightforward rotation of the frame" is close, but it's the total energy that is the time component of the four-momentum and mixes with the spatial momentum under Lorentz transformations. Kinetic energy is the difference between that transformed total energy and the invariant rest energy. So kinetic energy isn't itself a four-vector component, but it arises from how the time component changes when viewed from a different inertial frame.

ajross 13 hours ago [-]
To nitpick your nitpick: I know. But precision isn't the point here, it's to point out that there's an interesting and deeper symmetry at work. Energy and Momentum are not actually different quantities that vary in different ways but are still conserved via different laws. They're actually expressible as a single conserved vector quantity.

Details about the specifics were hidden behind the scare quotes on "rotation". But sure, my phrasing was loose, how about 'What we ses as "kinetic energy" pops out of the Lorentz "rotations" of that energy in different reference frames.' ...?

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